Problem: The following equation is true for all real values of $z$ for which the expression on the left is defined, and $B$ is a polynomial expression. $\dfrac{5z^2}{z^2+5z} \div \dfrac{B}{z^2-2z-35}=1$ What is $B$ ? $B=$
Answer: The left side of the equation is a quotient of two rational expressions, while the right side is simply $1$. This means that the numerator and the denominator of the resulting quotient on the left side should cancel out completely. In order to solve for $B$, let's divide the expressions and simplify as much as we can. We start with factoring the numerators and the denominators of the two expressions. [Why are we doing this?] The numerator, $5z^2$, of the first expression cannot be factored any further. The denominator, $z^2+5z$, of the first expression can be factored as $z(z+5)$ by factoring out a $z$. The denominator, $z^2-2z-35$, of the second expression can be factored as $(z+5)(z-7)$ using the sum-product pattern. Now the quotient looks as follows: $\dfrac{5z^2}{z(z+5)} \div \dfrac{B}{(z+5)(z-7)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=} \dfrac{5z^2}{z(z+5)} \div\dfrac{B}{(z+5)(z-7)} \\\\\\ &=\dfrac{5z^2}{z(z+5)} \cdot \dfrac{(z+5)(z-7)}{B} &\text{Flip the divisor} \\\\\\ &= \dfrac{5z^2(z+5)(z-7)}{z(z+5)B} &\text{Multiply across.}\\\\\\ &= \dfrac{5\cdot {\cancel{z}} \cdot z {\cancel{(z+5)}}(z-7)}{{\cancel{z}}{\cancel{(z+5)}}B} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{5z(z-7)}{B} \end{aligned}$ After this simplification, our equation now looks like: $\dfrac{5z(z-7)}{B}=1$ We can conclude that the numerator and the denominator of the expression on the left must be equal. [Why?] In other words, $B=5z(z-7)$, which is equivalent to $5z^2-35z$.